To find the time for one revolution (period) and the radius of the circle of a conical pendulum, we use the given information: • Length of the string, $L = 100 \text{ cm} = 1.0 \text{ m}$ • Angle with the vertical, $\theta = 30^\circ$ • We will use the acceleration due to gravity, $g = 9.8 \text{ m/s}^2$. Step 1: Calculate the radius of the circle, $r$. The radius of the circular path is related to the length of the string and the angle $\theta$ by trigonometry. $$ r = L \sin\theta $$ Substitute the given values: $$ r = (1.0 \text{ m}) \sin(30^\circ) $$ $$ r = (1.0 \text{ m}) (0.5) $$ $$ r = 0.5 \text{ m} $$ Converting to centimeters: $$ r = 0.5 \text{ m} \times \frac{100 \text{ cm}}{1 \text{ m}} $$ $$ r = 50 \text{ cm} $$ Step 2: Calculate the time for one revolution (period), $T$. For a conical pendulum, the period $T$ is given by the formula: $$ T = 2\pi \sqrt{\frac{L \cos\theta}{g}} $$ Substitute the given values: $$ T = 2\pi \sqrt{\frac{(1.0 \text{ m}) \cos(30^\circ)}{9.8 \text{ m/s}^2}} $$ We know that $\cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.8660$. $$ T = 2\pi \sqrt{\frac{1.0 \text{ m} \times 0.8660}{9.8 \text{ m/s}^2}} $$ $$ T = 2\pi \sqrt{\frac{0.8660}{9.8}} $$ $$ T = 2\pi \sqrt{0.088367... \text{ s}^2} $$ $$ T = 2\pi (0.297266... \text{ s}) $$ $$ T \approx 1.8676 \text{ s} $$ The radius of the circle is $\boxed{50 \text{ cm}}$. The time for one revolution (period) is $\boxed{1.87 \text{ s}}$.